Question

Find the equation of the straight line which passes through the point (4,3) and is perpendicular to: 2x-3y+6=0​

Answer 1

We have to find the slope of the given line. We can solve its equation for y to get

`2x-3y+6=0\iff 3y=2x+6 \iff y = (2)/(3)y+2`

So, the slope of the given line is 2/3.

If m and m' are the slopes of two perpendicular lines, we have

`mm'=-1`

This means that the perpendicular slope is -3/2.

Now we use the point-slope formula

`y-y_0=m(x-x_0)`

where
`(x_0,y_0)` is the given point and m is the given slope, to find the equation of the required line:

`y-3=-(3)/(2)(x-4) \iff y = -(3)/(2)x+9`